How binary Random number generation upto n nodes ?

i was do this
clc;
clear all;
close all;
% Link is (n-1)*node/2
node = 3;
offset = 1;
L = (node-offset)*node/2;
c =2^L;
% Generate c binary values:
r = randi([0, 1], c,L);
disp(r);
output:
0 0 1
1 0 0
1 0 0
1 0 1
1 1 0
1 0 0
0 1 0
0 0 1
but i want
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
please help me how to generate binary random numbers up to n in order wise

1 Kommentar

Your current code generates random patterns. But what you want is not random at all ... it is an ordered list. Why do you write you want random output but then show a non-random ordered list as what you want?

Melden Sie sich an, um zu kommentieren.

 Akzeptierte Antwort

James Tursa
James Tursa am 15 Mär. 2021
Here is how to get your stated desired output, but this is not random at all:
r = dec2bin(0:(2^L-1)) - '0';

4 Kommentare

ankanna
ankanna am 15 Mär. 2021
I also try this code but upto 7 nodes the numbers. If I enter n=9, the will not come. I want upto n nodes.
James Tursa
James Tursa am 15 Mär. 2021
Bearbeitet: James Tursa am 15 Mär. 2021
The memory requirements for generating the complete list of patterns can easily exceed the amount of memory that your computer has. This method is only practical for small numbers. I could suggest that you use logical variables instead of double, but even then if you have such a large list how do you intend to use this downstream in your code? Depending on how involved your downstream processing is, it might take days or weeks or months or years to process all of the individual patterns. You will need a different approach to solving your problem if you have to work with large numbers.
ankanna
ankanna am 24 Apr. 2021
n = 3;
Link=(n*(n-1))/2;
c=2^Link;
NN = toeplitz(Link+1:-1:2)
mask = logical(fliplr(diag(ones(1,Link-1),-1)));
NN(mask) = 1;
for c = 0:2^Link-1
l = bitget(c, NN)
end
the above code i generate all configuration matrix.
i need to generate all paths in this network.
please help me to generate all paths in the network
Jan
Jan am 24 Apr. 2021
This is a new question. Please ask it in a new thread.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Produkte

Version

R2016a

Gefragt:

am 15 Mär. 2021

Kommentiert:

Jan
am 24 Apr. 2021

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by